Find lim x→0 sin2x/sin3x | Evaluate lim x→0 sin3x/sin2x

The limit of sin2x/sin3x when x approaches 0 is equal to 2/3, that is, lim_x→0 sin2x/sin3x = 2/3. The limit of sin3x/sin2x when x approaches 0 is equal to 3/2, that is, lim_x→0 sin3x/sin2x = 3/2.

The following formula will be useful to compute the above limits:

lim_x→0 sinx/x = 1 …(∗)

Limit of sin2x/sin3x when x approaches 0

The limit of sin2x/sin3x when x→0 can be computed as follows:

lim_x→0 $\dfrac{\sin 2x}{\sin 3x}$

= lim_x→0 $\dfrac{\frac{\sin 2x}{2x}}{\frac{\sin 3x}{3x}} \times \dfrac{2}{3}$

= $\dfrac{2}{3}$ lim_x→0 $\dfrac{\frac{\sin 2x}{2x}}{\frac{\sin 3x}{3x}}$

By the product rule of limits, the above limit

= $\dfrac{2}{3}$ lim_x→0 $\frac{\sin 2x}{2x}$ lim_x→0 $\dfrac{1}{\frac{\sin 3x}{3x}}$

Let z=2x and t=3x. Then both z, t →0 when x→0.

= $\dfrac{2}{3}$ lim_z→0 $\frac{\sin z}{z}$ lim_t→0 $\dfrac{1}{\frac{\sin t}{t}}$

= $\dfrac{2}{3}$ × 1 × $\dfrac{1}{1}$ by the above limit rule (∗).

= 2/3

Therefore, the limit of sin2x/sin3x when x→0 is 2/3.

ALSO READ:

limx→0 (ex-1)/x = 1	limx→∞ sinx/x = 0
limx→0 x/cosx = 0	limx→0 x/sinx = 1

Limit of sin3x/sin2x when x approaches 0

We evaluate the limit x→0 sin3x/sin2x as follows:

lim_x→0 $\dfrac{\sin 3x}{\sin 2x}$

= lim_x→0 $\dfrac{\frac{\sin 3x}{3x}}{\frac{\sin 2x}{2x}} \times \dfrac{3}{2}$

= $\dfrac{3}{2}$ lim_x→0 $\dfrac{\frac{\sin 3x}{3x}}{\frac{\sin 2x}{2x}}$

By the quotient rule of limits, we have the above limit

= $\dfrac{3}{2}$ $\dfrac{\lim\limits_{x \to 0}\frac{\sin 3x}{3x}}{\lim\limits_{x \to 0} \frac{\sin 2x}{2x}}$

Taking z=2x and t=3x as before, we have

= $\dfrac{3}{2}$ $\dfrac{\lim\limits_{t \to 0}\frac{\sin t}{t}}{\lim\limits_{z \to 0} \frac{\sin z}{z}}$

= $\dfrac{3}{2}$ × $\dfrac{1}{1}$ by the above limit formula (∗).

= 3/2

So the limit of sin3x/sin2x when x→0 is 3/2.

Have You Read These Limits?

lim_x→0 tanx/x = 1

lim_x→∞ tanx/x = undefined

lim_x→0 sin(1/x) = undefined

Epsilon – delta definition of limit

FAQs